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  1. #1
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    pole

    Let R(z) be a rational function: R(z)=p(z)/q(z) where p and q are holomorphic polynomials. Let f be holomorphic on C\ \{P_1, P_2,....,P_k \} and suppose f has a pole at each of the points P_1, P_2,....,P_k. Finally assume that |f(z)| \leq |R(z)| for all z at which these functions are defined. Prove that f is a constant multiple of R.
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  2. #2
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    Quote Originally Posted by Kat-M View Post
    Let R(z) be a rational function: R(z)=p(z)/q(z) where p and q are holomorphic polynomials. Let f be holomorphic on C\ \{P_1, P_2,....,P_k \} and suppose f has a pole at each of the points P_1, P_2,....,P_k. Finally assume that |f(z)| \leq |R(z)| for all z at which these functions are defined. Prove that f is a constant multiple of R.

    You meant, I presume, that \{P_1,\ldots,P_k\} are the zeros of the denominator polynomial q(z) , right? Well, we get that q(z)f(z) is an entire function and

    |q(z)f(z)|\leq |p(z)|\Longrightarrow as no entire function can dominate another entire function unless one is a scalar multiple of the other (this follows at once from

    Liouville's Theorem), we get what we want.

    Tonio
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